Question -
Answer -
(i) 2 + 5 + 8 + … + 182
First term, a = a1 =2
Common difference, d =a2 – a1 = 5 – 2 = 3
an termof given AP is 182
an = a+ (n-1) d
182 = 2 + (n-1) 3
182 = 2 + 3n – 3
182 = 3n – 1
3n = 182 + 1
n = 183/3
= 61
Now,
By using the formula,
S = n/2 (a + l)
= 61/2 (2 + 182)
= 61/2 (184)
= 61 (92)
= 5612
∴ The sum of the seriesis 5612
(ii) 101 + 99 + 97 + … + 47
First term, a = a1 =101
Common difference, d =a2 – a1 = 99 – 101 = -2
an termof given AP is 47
an = a+ (n-1) d
47 = 101 + (n-1)(-2)
47 = 101 – 2n + 2
2n = 103 – 47
2n = 56
n = 56/2 = 28
Then,
S = n/2 (a + l)
= 28/2 (101 + 47)
= 28/2 (148)
= 14 (148)
= 2072
∴ The sum of the seriesis 2072
(iii) (a – b)2 +(a2 + b2) + (a + b)2 + s…. + [(a +b)2 + 6ab]
First term, a = a1 =(a-b)2
Common difference, d =a2 – a1 = (a2 + b2)– (a – b)2 = 2ab
an termof given AP is [(a + b)2 + 6ab]
an = a+ (n-1) d
[(a +b)2 + 6ab] = (a-b)2 + (n-1)2ab
a2 + b2 +2ab + 6ab = a2 + b2 – 2ab + 2abn – 2ab
a2 + b2 +8ab – a2 – b2 + 2ab + 2ab = 2abn
12ab = 2abn
n = 12ab / 2ab
= 6
Then,
S = n/2 (a + l)
= 6/2 ((a-b)2 +[(a + b)2 + 6ab])
= 3 (a2 +b2 – 2ab + a2 + b2 + 2ab + 6ab)
= 3 (2a2 +2b2 + 6ab)
= 3 × 2 (a2 +b2 + 3ab)
= 6 (a2 +b2 + 3ab)
∴ The sum of the seriesis 6 (a2 + b2 + 3ab)